Let $$y^{2}=12x$$ be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that $$\angle OPA =90^\circ$$. Then the locus of the centroid of such triangles OPA is:

We begin with the parabola $$y^2 = 12x$$ whose vertex is at O(0,0), and we note that point $$P$$ lies on the parabola, point $$A$$ lies on the x-axis, and $$\angle OPA = 90°$$.

We introduce the parametric form $$P = (3t^2, 6t)$$ (with $$a = 3$$) and set $$A = (h, 0)$$.

The condition $$\angle OPA = 90°$$ implies $$\vec{PO} \perp \vec{PA}$$, so we compute $$\vec{PO} = (-3t^2, -6t)$$ and $$\vec{PA} = (h - 3t^2, -6t)$$.

Orthogonality gives $$\vec{PO} \cdot \vec{PA} = 0$$, namely

$$-3t^2(h - 3t^2) + (-6t)(-6t) = 0$$

which simplifies to $$-3t^2 h + 9t^4 + 36t^2 = 0$$. Dividing by $$3t^2$$ (with $$t \neq 0$$) yields $$-h + 3t^2 + 12 = 0$$ and hence $$h = 3t^2 + 12$$.

Next, the centroid of triangle OPA is given by $$G = \Bigl(\frac{0 + 3t^2 + h}{3},\,\frac{0 + 6t + 0}{3}\Bigr).$$ Setting $$G = (X, Y)$$ leads to

$$X = \frac{3t^2 + h}{3} = \frac{3t^2 + 3t^2 + 12}{3} = \frac{6t^2 + 12}{3} = 2t^2 + 4,$$

$$Y = 2t\quad\Longrightarrow\quad t = Y/2.$$

Substitution into $$X = 2t^2 + 4$$ gives $$X = 2 \cdot \frac{Y^2}{4} + 4 = \frac{Y^2}{2} + 4,$$ so $$Y^2 = 2X - 8 = 2(X - 4)\,. $$

In standard variables this becomes $$y^2 = 2(x - 4)$$, i.e.\ $$y^2 - 2x + 8 = 0\,. $$

The answer is Option 2: $$y^2 - 2x + 8 = 0$$.